What volume of 0.130 M NaOH is needed to neutralize 4.00 g of benzoic acid?
moles benzoic acid = grams/molar mass
moles NaOH = moles benzoic acid
M = moles/L. You know M and moles. Calculate L.
To determine the volume of 0.130 M NaOH needed to neutralize 4.00 g of benzoic acid, we need to follow a series of steps.
Step 1: Write a balanced equation for the reaction between benzoic acid (C6H5COOH) and sodium hydroxide (NaOH). The reaction can be represented as follows:
C6H5COOH + NaOH -> C6H5COONa + H2O
Step 2: Calculate the molar mass of benzoic acid. The molar mass of benzoic acid (C6H5COOH) is determined by adding up the atomic masses of each element present in the compound:
C6H5COOH: (6 * atomic mass of C) + (5 * atomic mass of H) + (2 * atomic mass of O) = 12.01 * 7 + 1.01 * 7 + 16.00 * 2 = 122.13 g/mol
Step 3: Calculate the number of moles of benzoic acid using the formula:
moles = mass / molar mass
moles of benzoic acid = 4.00 g / 122.13 g/mol = 0.0327 mol
Step 4: Using the balanced equation, we can see that the stoichiometric ratio between benzoic acid (C6H5COOH) and sodium hydroxide (NaOH) is 1:1. This means that for every 1 mol of benzoic acid, we will need 1 mol of NaOH.
Since the concentration of NaOH is given as 0.130 M (moles per liter), we can calculate the volume of NaOH required using the formula:
volume = moles / concentration
volume of NaOH = 0.0327 mol / 0.130 mol/L = 0.251 L = 251 mL
Therefore, the volume of 0.130 M NaOH needed to neutralize 4.00 g of benzoic acid is 251 mL.